Problem: If the roots of the quadratic equation $\frac32x^2+11x+c=0$ are $x=\frac{-11\pm\sqrt{7}}{3}$, then what is the value of $c$?
By the quadratic formula, the roots of the equation are $$x=\frac{-(11)\pm\sqrt{(11)^2-4(\frac32)c}}{2(\frac32)},$$which simplifies to $$x=\frac{-11\pm\sqrt{121-6c}}{3}.$$This looks exactly like our target, except that we have to get the $121-6c$ under the square root to equal $7$. So, we solve the equation $121-6c=7$, which yields $c=\boxed{19}$.